Question

A $40\, \mu F$ capacitor in a defibrillator is charged to $3000 \,V$. The energy stored in the capacitor is sent through the patient during a pulse of duration $2\, ms$. The power delivered to the patient is

• A. 45 kW
• B. 90 kW
• C. 180 kW
• D. 360 kW

Answer 1

Answer: (B)

90 kW

Answer 2

A capacitor is a device that stores energy in the electric field created between a pair of conductors on which equal but opposite electric charges have been placed. The energy stored in a capacitor $=\frac{1}{2} C V^{2}$ Given, $C=40\, \mu F =40 \times 10^{-6} F , V=3000\, V$ $\therefore E=\frac{1}{2} \times 40 \times 10^{-6} \times(3000)^{2}$ $=180\, J$ Also $1 \,W =1 \,J / s$ $\therefore 2\, ms =2 \times 10^{-3} s$ Hence, power $=\frac{180 J }{2 \times 10^{-3} s }$ $=90 \times 10^{3} W =90\, kW$